Problem: Simplify; express your answer in exponential form. Assume $k\neq 0, y\neq 0$. $\dfrac{{(k^{3})^{-3}}}{{(k^{-3}y^{-1})^{-2}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{3}}$ to the exponent ${-3}$ . Now ${3 \times -3 = -9}$ , so ${(k^{3})^{-3} = k^{-9}}$ In the denominator, we can use the distributive property of exponents. ${(k^{-3}y^{-1})^{-2} = (k^{-3})^{-2}(y^{-1})^{-2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{3})^{-3}}}{{(k^{-3}y^{-1})^{-2}}} = \dfrac{{k^{-9}}}{{k^{6}y^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-9}}}{{k^{6}y^{2}}} = \dfrac{{k^{-9}}}{{k^{6}}} \cdot \dfrac{{1}}{{y^{2}}} = k^{{-9} - {6}} \cdot y^{- {2}} = k^{-15}y^{-2}$.